pop 1218 THE DRUNK JAILER java版

原题如下：
THE DRUNK JAILER
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 20311 Accepted: 12822
Description

A certain prison contains a long hall of n cells, each right next to each other. Each cell has a prisoner in it, and each cell is locked.
One night, the jailer gets bored and decides to play a game. For round 1 of the game, he takes a drink of whiskey,and then runs down the hall unlocking each cell. For round 2, he takes a drink of whiskey, and then runs down the
hall locking every other cell (cells 2, 4, 6, ?). For round 3, he takes a drink of whiskey, and then runs down the hall. He visits every third cell (cells 3, 6, 9, ?). If the cell is locked, he unlocks it; if it is unlocked, he locks it. He
repeats this for n rounds, takes a final drink, and passes out.
Some number of prisoners, possibly zero, realizes that their cells are unlocked and the jailer is incapacitated. They immediately escape.
Given the number of cells, determine how many prisoners escape jail.
Input

The first line of input contains a single positive integer. This is the number of lines that follow. Each of the following lines contains a single integer between 5 and 100, inclusive, which is the number of cells n.
Output

For each line, you must print out the number of prisoners that escape when the prison has n cells.
Sample Input

2
5
100
Sample Output

2
10

解法：求1-n之间有多少个平方数

我的java代码：

import java.util.Scanner;

public class Main{
	public static void main(String args[])
	{
		Scanner cin = new Scanner(System.in);
		int numberNum = cin.nextInt();
		int numbers[] = new int[numberNum];
		for(int i = 0; i < numberNum; i++){
			numbers[i] = cin.nextInt();
		}
		
		for(int k = 0; k < numberNum; k++)
		{
			int unlockedNum = 0;
			
			for(int i = 1; i <= numbers[k]; i++)
			{
				for(int j = 1; j < numbers[k]; j++)
				{
					if(i == j*j)
					{ //if i is the square of j,then the door at the place is unlocked
						unlockedNum++;
						break;
					}
				}
			}
			System.out.println(unlockedNum);
		}
	
	}
	
}

望同志们多多指教！

评论

2 楼 softliumin 2013-07-17  

你加入不是模拟过程，而是通过里面的规律开结题，那么可以这么写：
Scanner cin = new Scanner(System.in);

int n = cin.nextInt();
for (int i = 0; i < n; i++)
{
System.out.println((int) Math.sqrt((double) cin.nextInt())  );
}
cin.close();

1 楼 softliumin 2013-07-17  

这个代码，我看了好久，才看懂······